Problem: Let $R$ be the region enclosed by the polar curve $r(\theta)=2-2\cos(\theta)$ where $\dfrac{2\pi}{3}\leq \theta\leq \pi$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^{\scriptsize\dfrac{2\pi}{3}}\left(2-4\cos(\theta)+2\cos^2(\theta)\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}\left(1-\cos(\theta)\right)d\theta$ (Choice C) C $ \int_0^{\scriptsize\dfrac{2\pi}{3}}\left(1-\cos(\theta)\right)d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}\left(2-4\cos(\theta)+2\cos^2(\theta)\right)d\theta$
Solution: This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=2-2\cos(\theta)}$, ${\alpha=\dfrac{2\pi}{3}}$, and ${\beta=\pi}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{2\pi}{3}}}^{{\pi}}\dfrac{1}{2}\left({2-2\cos(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}\dfrac{1}{2}\left(4-8\cos(\theta)+4\cos^2(\theta)\right)d\theta \\\\ &= \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}\left(2-4\cos(\theta)+2\cos^2(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{2\pi}{3}}^{\pi}\left(2-4\cos(\theta)+2\cos^2(\theta)\right)d\theta$